Basic explanation of Special Relativity.

1. S S’
v
Now, suppose that we have two similar spacecraft, denoted as S and S’. S’ is moving relative to S at a constant velocity, v.
S
S’
v
Position A

2. S’
S
v
X
X’
S said:
“The light flash occurred when I
was x distance from it and when
my clock is marked as t.”
S’ said:
“The light flash occurred when I
was x’ distance from it and when
my clock is marked as t’.”
Position A Position B

3. S
x = x’ + vt’
S said:
“The light travels in distance x in
time t, measured on my clock.
Thus, c = x/t”
S’ said:
“The light travels in distance x’ in
time t’, measured on my clock.
Thus, c = x’/t’ ”
t'
t
Position A Position B
S’
v
x = x’ + the separation between S and S’, or the distance between point A
and point B. Relative velocity, v multiplied by the time that has elapsed,
which is, depending on which clock you are measuring, t or t’.
x’ = x - vt
We know that speed of light, c is a constant

4. We know that speed of light, c is a constant. So as x ≠ x’,
thus t ≠ t’.
Each of them alleges that the other has made an error in
measuring distance, time or both. They assign an error called
gamma, ϒ.
S’ says:
x’=ϒ(x - vt)
S says:
x=ϒ(x’ + vt’)
xx’ = ϒ2
(x’ + vt’)(x - vt)
xx’ = ϒ2
(x’x + vt’x – x’vt - 𝑣2t’t)
From c = x/t or c = x’/t’,
t = x/c or t’ = x’/c
xx’ = ϒ2
[x’x + vx(x’/c) – x’v(x/c) - 𝑣2
(x’/c)(x/c)]
Divide by xx’, we get:
1 = ϒ2
[1 + (v/c) – (v/c) – (𝑣2
/𝑐2
)]
ϒ = 1/ (1 − 𝑣2/𝑐2
)
Substituting the value of ϒ into x and x’, we get:
x = x’ + vt’/ (1 − 𝑣2/𝑐2
) and x’ = x - vt/ (1 − 𝑣2/𝑐2
)
The gamma term is only relevant when v is approaching c. If v is too small, the
portion of the square root reduces to 1, and it goes back to Newtonian
Mechanics.
Lorentz Factor
From ① and ②,
x=ϒ(x’ + vt’) x’=ϒ(x - vt)
x=ϒ(x’ + v(x’/c) x’=ϒ(x – v(x/c))
Divide by c,
x/c=ϒ(x’/c + v(x’/𝑐2
)
x’/c=ϒ(x/c – v(x/𝑐2))
Then,
t = ϒ[t’ + (vx’/𝑐2
)]
t’ = ϒ[t – (vx/𝑐2
)]
①
②
③
④

5. TIME DILATION
1m
Imagine that we make a very simple clock. It is made up of two mirror
facing each other. Then there is a light beam that travels from the bottom
mirror to top mirror. When the light beam hits the lower mirror, it hits the
counter which clicks on by 1 count.
We know that c = 3𝑥108
m/s and the distance from the bottom mirror to
top mirror to lower mirror = 2m.
Light beam
mirror
Now imagine this simple clock is onboard spacecraft S
v
S
v v

6. 1m
The observer sees that the light
beam has travelled further than
2m, because it moves at an angle,
ϴ
ϴ
Since the speed of light is invariant, it must have taken
longer for the light beam to get up to top mirror than
the person onboard the spacecraft thinks.
This leads to the conclusion that comes from the equation ④ which tells, a clock
moving relative to you will run slow.
LENGTH CONTRACTION
L L’
Now suppose that we have two spacecraft, S and S’.
There is a rod, called L’ in spacecraft S’. The person in
spacecraft S measures the same rod as L, relative to
his position.
Using formula ②, S’ said:
L’ = L – 0 / (1 − 𝑣2/𝑐2
)
So,
L = L’ (1 − 𝑣2/𝑐2)
Because there is no
difference in time
A moving rod appears to
be shorter. Length
contracts.
t’ = ϒ[t – (vx/𝑐2)]
x’ = x - vt/ (1 − 𝑣2/𝑐2
)